Let for a differentiable function f : (0, ∞) → R, f(x) - f(y) ≥ log_e(x/y) + x - y, ∀x, y ∈ (0, ∞).

Q: Why do we have to check both $x > y$ and $x < y$?

Great question! When we divide the inequality $f(x) - f(y) \ge g(x) - g(y)$ by $(x - y)$, the direction of the inequality depends on the sign of $(x - y)$. If $x > y$, $(x - y)$ is positive, so the sign stays the same. If $x < y$, $(x - y)$ is negative, so the sign flips. We need both cases to 'squeeze' the derivative from both sides, proving that $f'(x)$ must equal $g'(x)$.

Q: How did you know to define $g(x) = \ln x + x$?

It is all about pattern recognition! The term $\ln(x/y) + x - y$ can be split into $(\ln x + x) - (\ln y + y)$ using logarithmic properties. By grouping the $x$ terms and $y$ terms, we reveal a function $g(x)$ that is being compared to $f(x)$. This is a classic technique in functional equations to isolate the behavior of the function.

Q: Is it always true that if $f'(x) \ge g'(x)$ and $f'(x) \le g'(x)$, then $f'(x) = g'(x)$?

Yes, absolutely! This is a direct consequence of the properties of inequalities. If a value is both greater than or equal to and less than or equal to another value, it must be equal to it. It is essentially the 'Squeeze Theorem' applied to the derivative definition.

Question

Let for a differentiable function $f : (0, \infty) 	o R, f(x) - f(y) \ge \log_e(\frac{x}{y}) + x - y, \forall x, y \in (0, \infty)$. Then $\sum_{n=1}^{20} f'(\frac{1}{n^2})$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical realm. Today, we are going to peel back the layers of a problem that might look intimidating at first glance, but is actually a masterclass in elegance.

We are given a differentiable function $f : (0, \infty) 	o \mathbb{R}$ satisfying the inequality:

$$f(x) - f(y) \ge \ln(x/y) + x - y$$

At first, this looks like a chaotic mess of variables. But in mathematics, chaos is often just order waiting to be discovered.

### The Art of Grouping

The first step in our journey is to find the hidden symmetry. Look at the right-hand side of our inequality: $\ln(x/y) + x - y$.

Using the properties of logarithms, we know that $\ln(x/y) = \ln x - \ln y$. If we substitute this back into our inequality, we get:

$$f(x) - f(y) \ge \ln x - \ln y + x - y$$

Now, let us group the terms involving $x$ together and the terms involving $y$ together. We can rewrite this as:

$$f(x) - (\ln x + x) \ge f(y) - (\ln y + y)$$

This is the "Aha!" moment. If we define a new function $g(x) = \ln x + x$, our inequality simplifies beautifully to:

$$f(x) - g(x) \ge f(y) - g(y)$$

### The Geometry of the Derivative

Now, let us think about what this means geometrically. We are looking for the derivative of $f(x)$. Recall the definition of the derivative:

$$f'(x) = \lim_{y 	o x} \frac{f(x) - f(y)}{x - y}$$

To get there, let us divide our inequality $f(x) - f(y) \ge g(x) - g(y)$ by $(x - y)$.

If we assume $x > y$, then $(x - y)$ is positive, and the inequality sign remains unchanged:

$$\frac{f(x) - f(y)}{x - y} \ge \frac{g(x) - g(y)}{x - y}$$

As we take the limit as $y 	o x$, the secant line becomes the tangent line, and we get $f'(x) \ge g'(x)$.

### The Trap and the Squeeze

Here is where many students stumble. We must also consider the case where $x < y$. In this scenario, $(x - y)$ is negative.

When we divide by a negative number, the inequality sign flips! So:

$$\frac{f(x) - f(y)}{x - y} \le \frac{g(x) - g(y)}{x - y}$$

Taking the limit as $y 	o x$ again, we find $f'(x) \le g'(x)$.

Now, we have our squeeze: $f'(x) \ge g'(x)$ and $f'(x) \le g'(x)$. The only logical conclusion is that $f'(x) = g'(x)$.

Since $g(x) = \ln x + x$, its derivative is simply $g'(x) = \frac{1}{x} + 1$. Thus, we have:

$$f'(x) = \frac{1}{x} + 1$$

### The Summation: Bringing it Home

We have conquered the functional part. Now, we must evaluate the sum $\sum_{n=1}^{20} f'(\frac{1}{n^2})$.

Substituting $x = \frac{1}{n^2}$ into our derivative formula, we get:

$$f'\left(\frac{1}{n^2}ight) = \frac{1}{1/n^2} + 1 = n^2 + 1$$

Our sum becomes $\sum_{n=1}^{20} (n^2 + 1)$. Using the linearity of summation, we split this into:

$$\sum_{n=1}^{20} n^2 + \sum_{n=1}^{20} 1$$

The sum of the first $n$ squares is given by the formula $\frac{n(n+1)(2n+1)}{6}$. For $n=20$, this is:

$$\frac{20 	imes 21 	imes 41}{6} = 2870$$

Adding the sum of the constant $1$ twenty times, which is $20$, we arrive at our final answer:

**$2870 + 20 = 2890$**

Animated Solution for Mathematics - Differentiation: Let for a differentiable function $f : (0, \infty) \to R, f (x) - f (y) \geq lo g_{e} (\frac{x}{y}) + x - y, \forall x, y \in (0, \infty)$ . Then $\sum_{n = 1}^{20} f^{'} (\frac{1}{n ^{2}})$ is equal to

Enter Numerical Value:

Visualized Solution

The Functional Inequality

Rearranging the Terms

Case 1: $x > y$ and the Secant

Applying the Limit ( $y \to x$ )

Case 2: $x < y$ (The Trap)

Limit for Case 2

Exact Derivative of $f (x)$

Evaluating the General Term

Setting up the Summation

Applying Summation Formulas

Final Calculation

The Sigma Insight: Differentiability of a Function

Analyzing the Setup

The Art of Grouping

The Geometry of the Derivative

The Trap and the Squeeze

The Summation

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